Perturbed Divisible Sandpiles and Quadrature Surfaces

Abstract

The main purpose of the present paper is to establish a link between quadrature surfaces (potential theoretic concept) and sandpile dynamics (Laplacian growth models). For this aim, we introduce a new model of Laplacian growth on the lattice $\mathbb {Z}^{d}$ (d ≥ 2) which continuously deforms occupied regions of the divisible sandpile model of Levine and Peres (J. Anal. Math. 111(1), 151–219 2010), by redistributing the total mass of the system onto $\frac 1m$-sub-level sets of the odometer which is a function counting total emissions of mass from lattice vertices. In free boundary terminology this goes in parallel with singular perturbation, which is known to converge to a Bernoulli type free boundary. We prove that models, generated from a single source, have a scaling limit, if the threshold m is fixed. Moreover, this limit is a ball, and the entire mass of the system is being redistributed onto an annular ring of thickness $\frac 1m$. By compactness argument we show that when m tends to infinity sufficiently slowly with respect to the scale of the model, then in this case also there is scaling limit which is a ball, with the mass of the system being uniformly distributed onto the boundary of that ball, and hence we recover a quadrature surface in this case. Depending on the speed of decay of 1/m, the visited set of the sandpile interpolates between spherical and polygonal shapes. Finding a precise characterisation of this shape-transition phenomenon seems to be a considerable challenge, which we cannot address at this moment.

Appendix: Uniqueness of the Scaling Limit

For any m,k > 0 fixed, let u (being compactly supported)¹ solve

$$ {\Delta} u = m \mathbb{I}_{\{0<u< k\}} - \delta_{0} . $$

(A.1)

We want to prove that such a u is unique and spherically symmetric. To see this, we show that if u is not spherically symmetric, then we may generate two distinct spherically symmetric solutions by starting from infimum/supremum over all rotations of u. Then, we prove that given the solution is spherically symmetric, then it is uniquely determined by coefficients m and k. These two components put together settle the uniqueness of solutions to Eq. A.1.

We shall now take the smallest super-solution u_∗ and we only need to check that the infimum in the class of super-solutions does not degenerate to zero (i.e. u_∗≢0) and that it solves the problem. Obviously we may only consider spherically symmetric super-solutions with shrinking radius of support. Hence integration by parts (using that u_∗ vanishes outside a large ball) implies Volume({0 < u_∗ < k}) ≥ 1/m. This bound, coupled with the spherical symmetry of u_∗ implies that the support of u_∗ must contain a ball of a fixed radius, in particular the support cannot shrink too much. So it remains to show that the smallest super-solution is actually a solution. Now in the set {u_∗ > k}, we may always replace u_∗ by h which is the solution of Δh = −δ₀, with boundary values k on ∂{u_∗ > k}. Also in the set {u_∗ < k} we make a replacement with solution to ${\Delta } h = k\mathbb {I}_{\{h>0 \}}$ and the corresponding boundary values. Hence we may assume ${\Delta } u_{*} = m \mathbb {I}_{\{0<u< k\}} - \delta _{0} - \mu _{*}$ , where support(μ_∗) is on the sphere ∂{u_∗ > k}. Next let v_∗ solve ${\Delta } v_{*} = -\delta _{0} + m\mathbb {I}_{\{0< u_{*} < k\}}$ and v_∗ = 0 on the boundary of the support of u_∗. Then by comparison principle v_∗≤ u_∗, and hence ${\Delta } v_{*} \leq - \delta _{0} + m\mathbb {I}_{\{0< v_{*} < k\}}$, implying that v_∗ is also a super-solution to our problem. Now u_∗ being infimum in the class implies v_∗ = u_∗, and hence u_∗ solves (A.1).

A similar argument works by taking the largest sub-solution, among all solutions with bounded support. Here again we may consider spherically symmetric sub-solutions. We need to show that the largest sub-solution in the class stays bounded. As in previous case we may replace the maximizing sequence of sub-solutions $u_{j}^{*}$ with exact solutions in the sets $\{ u_{j}^{*} > k \}$ and $\{u_{j}^{*} < k \}$. To see that the support of such sub-solutions should stay uniformly bounded we argue as follows. Let $R_{j}^{*}$ be the radius of the support of $u^{*}_{j}$, and R_j be such that $u_{j}^{*} = k$ on |x| = R_j. Using sub-solutions’ properties, we can conclude that $Volume(\{ 0< u^{*}_{j} < k\}) \leq 1$ (this is the reverse of the previous inequality). Hence $R_{j}^{*} \approx R_{j} + c_{d}R_{j}^{1-d}$, for some dimensional constant c_d > 0.

Let now F denote the fundamental solution, and set $F_{j} =\max (F - a_{d}(R_{j}^{*})^{2-d}, 0 )$, for d ≥ 3 and $F_{j} = a_{2} \max (\log (R_{j}^{*}/|x|), 0 )$, where a_d is a normalization constant for d ≥ 2. Since

$${\Delta} F_{j} = -\delta_{0} \leq -\delta_{0} + m \mathbb{I}_{\{0< u^{*}_{j} < 1/k \}} \leq {\Delta} u^{*}_{j}, \qquad \text{in } B_{R_{j}^{*}} $$

and

$$F_{j} > 0 = u^{*}_{j} \quad \text{on } \partial B_{R^{*}_{j}},$$

we can apply comparison principle to deduce $F_{j} \geq u^{*}_{j}$ on $B_{R_{j}^{*}}$. But on the other hand for $|x| = R_{j} \approx R_{j}^{*} - c_{d}R_{j}^{1-d}$ we have

$$F_{j} (x) = aR_{j}^{2-d} - a(R_{j}^{*})^{2-d} < 1/k = u_{j}^{*} (x)$$

once $R_{j}^{*} \approx R_{j} + c_{d}R_{j}^{1-d}$ is large enough. Hence a contradiction. This implies that the support of sub-solutions should stay bounded. Now taking the largest sub-solution, and using the fact that their supports are uniformly bounded, we get yet another solution to Eq. A.1.

The conclusion is that if Eq. A.1 has a solution with bounded support, which is not spherically symmetric, then we can produce two distinct spherically symmetric solutions. Hence to show uniqueness of solutions to Eq. A.1 with bounded support, it suffices to show that there is only one spherically symmetric solution to Eq. A.1 having bounded support. This we establish in the next lemma.

Lemma A.1

For anym, k > 0 there is a unique spherically symmetric solution to Eq.A.1having bounded support.

Proof

We shall give a computational proof, which is elementary, but tedious. First we consider the case when d > 2. Let ω_d > 0 be the normalising constant of the Green’s kernel.² The spherical symmetry of u implies that it should be of the form

$$u(x) = \left\{\begin{array}{lllllll} a_{1} + \omega_{d} |x|^{2-d} , &\text{if} 0<|x|\leq r_{1}, \\ a_{2} + a_{3}|x|^{2-d} + \frac{m}{2d} |x|^{2} , &\text{if } r_{1}<|x|\leq r_{2}, \end{array}\right. $$

where a_i and r_j are constants. The aim is to show that these constant are uniquely determined from Eq. A.1, which we do next.

By a straightforward computation we have

$$ a_{2} + a_{3} r_{2}^{2-d} + \frac{m}{2d} {r_{2}^{2}} = 0 $$

(A.2)

$$ a_{3} (2-d) r_{2}^{-d} + \frac md = 0 $$

(A.3)

$$ a_{1} + \omega_{d} r_{1}^{2-d} = k $$

(A.4)

$$ a_{2} + a_{3} r_{1}^{2-d} +\frac{m}{2d} {r_{1}^{2}} = k $$

(A.5)

$$ \omega_{d} (2-d) r_{1}^{-d} = a_{3} (2-d) r_{1}^{-d} + \frac md, $$

(A.6)

where Eq. A.2 is due to the condition u = 0 on |x| = r₂, (A.3) comes from |∇u| = 0 on |x| = r₂, (A.4) is in view of u = k on |x| = r₁ and (A.5) is the continuity of u on |x| = r₁, finally (A.6) is the continuity of the gradient on |x| = r₁.

From Eqs. A.6 and A.3 we have

$$ a_{3}= \omega_{d} + \frac{m}{d(d-2)} {r_{1}^{d}} = \frac{m}{d(d-2)} {r_{2}^{d}}, $$

(A.7)

and hence

$$ r_{2}= \left[ {r_{1}^{d}} + \omega_{d} \frac{d (d-2)}{m} \right]^{1/d}. $$

(A.8)

Subtracting (A.3) from Eq. A.5, to eliminate a₂, we obtain

$$ \frac{m}{d(d-2)} r_{1}^{2-d} \left[ {r_{1}^{d}} + \omega_{d} \frac{d (d-2)}{m} \right] +\frac{m}{2d} {r_{1}^{d}} - \frac{m}{2(d-2)} {r_{2}^{2}} = k. $$

(A.9)

Plugging the value of r₂ from Eq. A.8, we get

$$ \frac{m}{2(d-2)}{r_{1}^{2}} + \omega_{d} r_{1}^{2-d} - \frac{m}{2(d-2)} \left[ {r_{1}^{d}} + \omega_{d} \frac{d(d-2)}{m} \right]^{\frac{2}{d}} =k. $$

(A.10)

To complete the proof we need to see that (A.10) has a unique solution when 0 < r₁ < ∞. Rearranging (A.10), consider the function

$$ f(x) = k - \frac{\omega_{d}}{x^{d-2}} - \frac{m}{2(d-2)} x^{2} + \frac{m}{2(d-2)} \left[ x^{d} + \omega_{d} \frac{d(d-2)}{m} \right]^{\frac 2d}, $$

(A.11)

where 0 < x < ∞. We need to show that f has a unique zero in (0,∞).

Observe, that f(0+) = −∞, and f( + ∞) = k > 0, and hence the continuity of f gives the existence of a zero for f. We are left with establishing the uniqueness of this zero. To this end, computing the derivative of f we get

$$\begin{array}{@{}rcl@{}} f^{\prime}(x) = (d-2) \frac{\omega_{d}}{x^{d-1}} + x \frac{m}{d-2} \left[ \left( 1+\frac{\omega_{d}}{x^{d}} \frac{d(d-2)}{m} \right)^{\frac{2-d}{d}} -1 \right] =\\ x \left\{ (d-2) \frac{\omega_{d}}{x^{d}} + \frac{m}{d-2} \left[ \left( 1+\frac{\omega_{d}}{x^{d}} \frac{d(d-2)}{m} \right)^{\frac{2-d}{d} } -1 \right] \right\} . \end{array} $$

We want to show that f^′(x) > 0 everywhere, which will complete the proof. Ignoring the x in front of f^′ in the last expression and denoting y := (d − 1)ω_dx^−d, we obtain that f^′(x) = xF(y), where

$$F(y) = y + \frac{m}{d-2} \left[ \left( 1+\frac dm y \right)^{\frac{2-d}{d}} -1 \right], $$

and hence it is enough to show that F > 0 everywhere. The latter follows easily from the fact that F(0) = 0 and F^′ > 0 in (0,∞). We conclude the proof of uniqueness in the case when d ≥ 3.

We now treat the case of d = 2. Set ω₂ = − 1/(2π) which is the normalising constant for the Green’s kernel. Again, the exact value of ω₂ is not important for the proof, and we will only use that ω₂ < 0. Relying on the spherical symmetry of u, as above, we see that u has to be of the form

$$u(x) = \left\{\begin{array}{lllllll} a_{1} + \omega_{2} \log |x| , &\text{if \(0<|x|\leq r_{1}\)}, \\ a_{2} + a_{3} \log|x| + \frac{m}{4} |x|^{2} , &\text{if } r_{1}<|x|\leq r_{2}, \end{array}\right. $$

where a_i and r_j are unknown constants. The goal is to show that these constants are being determined uniquely given the properties of u. As above, a direct computation leads to

$$ a_{2} + a_{3} \log r_{2} + \frac{m}{4} {r_{2}^{2}} = 0 $$

(A.12)

$$ a_{3} r_{2}^{-2} + \frac m2 = 0 $$

(A.13)

$$ a_{1} + \omega_{2} \log r_{1} = k $$

(A.14)

$$ a_{2} + a_{3} \log r_{1} + \frac{m}{4} {r_{1}^{2}} = k $$

(A.15)

$$ \omega_{2} r_{1}^{-2} = a_{3} r_{1}^{-2} + \frac m2, $$

(A.16)

where (A.12) is due to the condition u = 0 on |x| = r₂, Eq. A.13 comes from |∇u| = 0 on |x| = r₂, Eq. A.14 is in view of u = k on |x| = r₁ and (A.15) is the continuity of u on |x| = r₁, finally (A.16) is the continuity of the gradient on |x| = r₁.

Using (A.13) we get $a_{3}= -\frac m2 {r_{2}^{2}}$, which together with Eq. A.16 implies

$$r_{2}= \left( {r_{1}^{2}} -\frac 2m \omega_{2} \right)^{\frac 12}. $$

Subtracting (A.15) from (A.12) we obtain

$$a_{3} \log \frac{r_{1}}{r_{2}} + \frac m4 ({r_{1}^{2}} - {r_{2}^{2}} ) = k. $$

From the above relation between r₁ and r₂ we have

$$k - \frac{\omega_{2}}{2} = a_{3} \log \frac{r_{1}}{r_{2}} = - \frac{a_{3}}{2} \log \left( \frac{r_{2}}{r_{1}} \right)^{2} $$

Plugging the values of a₃ and r₂ we get

$$ k- \frac{\omega_{2}}{2} = \frac m4 \left( {r_{1}^{2}} - \frac 2m \omega_{2} \right) \log \left( 1- \frac 2m \frac{\omega_{2}}{{r_{1}^{2}}} \right), $$

(A.17)

and thus need to show that the last equation has a unique solution in (0,∞). To this end, set $y := -\frac {2}{m} \frac {\omega _{2}}{{r_{1}^{2}}}$. With this notation, we need to prove that the function

$$F(y) = -\frac 12 \omega_{2} \left( 1+ \frac 1y \right) \log (1+y) - k +\frac{\omega_{2}}{2}, $$

has a unique zero in the interval (0,∞). Observe, that $F(0+) = - k + \frac {\omega _{2}}{2}<0$ as ω₂ < 0, and F( + ∞) = + ∞, hence the existence. For uniqueness, computing the derivative of F, we see

$$F^{\prime}(y) = -\frac{\omega_{2}}{2y} \left( 1-\frac{\log(1+y)}{y} \right), $$

which is always positive in the range y ∈ (0,∞), and the uniqueness follows.

The proof of the lemma is now complete. □

As we saw in Section 4, the scaling limit of the sandpile generated by a single source at the origin, is determined by the following PDE

$$ {\Delta} u = 2d m \mathbb{I}_{\{ 0<u<1/m }\} - 2d \delta_{0} \text{ in } \mathbb{R}^{d}, $$

(A.18)

where the support of u is bounded and contains a ball of some fixed radius, in both cases uniformly with respect to m. From Lemma A.1 we have that u is spherically symmetric, moreover, if we let x_m be the radius of the ball $\{ u\leq \frac 1m \}$, then following (A.11) and (A.17) we have that x_m is the unique zero in the range (0,∞) of the function

$$ f(x) = 1 - 2d m \frac{\omega_{d} }{x^{d-2}} - m^{2} \frac{d}{d-2} x^{2} + m^{2} \frac{d}{d-2} \left[ x^{d} + \omega_{d} \frac{d(d-2)}{m} \right]^{\frac 2d}, $$

(A.19)

for dimension d > 2, and for dimension 2, rearranging (A.17), and plugging the value of $\omega _{2} = -\frac {1}{2\pi }$, we see that x_m is the unique solution in the range (0,∞) of the equation

$$ \frac{1}{m^{2}} + \frac{1}{\pi m} = \left( x^{2} + \frac{1}{\pi m} \right) \log \left( 1+\frac{1}{\pi m x^{2}} \right). $$

(A.20)

Our final result shows that the limiting shapes of the sandpiles, corresponding to each m converge, as m →∞. For this it will be enough to prove that the radii x_m converge. This we do next; the proof is elementary and straightforward.

Lemma A.2

The radiix_mconverge asm →∞.

Proof

Due to their relation with the sandpile, the set {x_m}_m≥ 1 is bounded away from zero and infinity, as we outlined above. We first consider the case when d > 2. Rearranging the last two terms in Eq. A.19 we get, for x = x_m, that

$$\begin{array}{@{}rcl@{}} f(x) = 1 - 2d m \frac{\omega_{d} }{x^{d-2}} + m^{2} \frac{d}{d-2} x^{2} \left[ \left( 1 + \frac{\omega_{d}}{x^{d}} \frac{d(d-2)}{m } \right)^{\frac 2d} -1 \right] = \\ 1 - 2d m \frac{\omega_{d} }{x^{d-2}} + m^{2} \frac{d}{d-2} x^{2} \left[ \frac 2d \frac{\omega_{d}}{x^{d}} \frac{d(d-2)}{m} + \frac 12 \frac 2d \left( \frac 2d -1 \right) {\omega_{d}^{2}} \frac{(d(d-2))^{2}}{x^{2d} m^{2}} + \mathrm{O}(m^{-3}) \right] = \\ 1- x^{2-2d} {\omega_{d}^{2}} (d-2)^{2} d + \mathrm{O}(m^{-1}), \text{ as } m\to \infty, \end{array} $$

where we have used the fact that {x_m} is bounded away from 0 and infinity. Since f(x_m) = 0, from the last expression, taking m →∞, we see that

$$\lim\limits_{m\to \infty} x_{m} = (d(d-2)^{2} {\omega_{d}^{2}} )^{\frac{1}{2(d-2)}}. $$

We now settle the case of d = 2. Defining y = πmx² in Eq. A.20, we get

$$1+\frac{\pi}{m} = (y + 1) \log \left( 1+\frac 1y \right). $$

Now, using that x_m, the solution to Eq. A.20 is bounded from 0 and infinity, we do asymptotic expansion in the last expression, when m →∞ (and hence y →∞), which gives

$$1+\frac{\pi}{m} = 1+ \frac{1}{2y} + \mathrm{O} (y^{-2}) \text{ as } m\to \infty, $$

and hence, as $y=\pi m {x_{m}^{2}}$, we obtain that

$$ \lim\limits_{m\to\infty} x_{m} = \frac{1}{\sqrt{2}} \frac{1}{\pi}. $$

(A.21)

The proof of the lemma is now complete. □